Discussion
A problem has 20 variables each with 5 values. How large is the assignment space?
Answer: D
Each variable independently takes any of its values, so the count is:
\[ 5^{20} \approx 9.5 \times 10^{13} \]
which is 5 to the 20. That figure is why enumeration is hopeless and why propagation matters so much. Computing the size of the space before choosing a method takes a moment and immediately rules out the naive approaches, which is exactly the estimate that should precede any implementation.
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